document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1018108>Question 1018108</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello could you please help me simplify: log little3 3/4+ log little3 4/5+ log little3 5/6+ log little3 6/7+ log little3 7/8+ log little3 8/9 / log 4/5+ log 5/6+ log 6/7+ log 7/8+ log 8/9+ log 9/10 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #634277 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=634277\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log 3 (3/4)(4/5)...(8/9))---everything cancels except for 3/9, or 1/3, the numerator.<BR>\n" );
document.write( "denominator log 3(4/5(5/6)...(9/10) or 4/10 or 2/5.<BR>\n" );
document.write( "This is log 3 (1/3)/log 3 (2/5)<BR>\n" );
document.write( "log 3 (1/3)=log 3 (1)- log 3 (3)=0-1 or -1.<BR>\n" );
document.write( "log 3 2/5 is log 3 2/log 3/5<BR>\n" );
document.write( "you are dividing by the last, so you invert and multiply by -1.<BR>\n" );
document.write( "-log 3(5)/log 3 (2)<BR>\n" );
document.write( "That is log 3 (2)/log 3(5), or log 3 (2/5)<BR>\n" );
document.write( "Notice that -log3 (1/3) is -(log3 1-log3 3)= -(-1) or 1.  The minus sign inverts the log.<BR>\n" );
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